Departures from Linearity

Unit 2 · Exploring Two-Variable Data

What AP Stats asks here

When the residual plot or scatterplot reveals curvature, the linear model is wrong and a transformation may rescue it. The trio of vocabulary words — leverage (unusual $x$ ), y-outlier (large residual), influential (both) — describes how single points affect the fit. AP rewards principled handling: never silently drop points; report results with and without.

Transformations

Exponential growth

\log y \text{ vs } x \;\;\text{linearizes}

Leveling-off (asymptotic) growth

y \text{ vs } \log x \;\;\text{linearizes}

Power

y = a x^k

\log y \text{ vs } \log x \;\;\text{linearizes}

Point taxonomy

Leverage

\text{unusual } x

y-outlier

\text{large residual}

Influence

\text{ability to move the fit; max when both above}

AP Tip: Diagnostic workflow: plot the scatterplot, fit the LSRL, plot the residuals, transform if there is structure, refit, recheck. Iterate until residuals look like random noise.

Caution: Dropping an outlier just to make r prettier is data manipulation. Defensible removals require a documented recording error or a different-population argument.

Type 1

Transformation choice

Match the visible curvature to a transformation. Exponential growth ⇒ $\log y$ . Leveling-off growth ⇒ $\log x$ . Multiplicative spread ⇒ $\log y$ stabilizes variance.

Example 1

A scatterplot of bacteria count over time curves sharply upward, and the residual plot from the LSRL is U-shaped. Which transformation is most appropriate? (A) No transformation — accept the LSRL because the points are tightly clustered. (B) Transform

y

via

\log y

. Bacteria growth is multiplicative/exponential, so

\log y

vs time is approximately linear. (C) Transform

x

via

\log x

. Time should always be log-transformed. (D) Use the LSRL but extrapolate only to early times.

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Type 2

Leverage, y-outliers, influence

Leverage describes unusual $x$ . y-outlier describes large residual. A point that is both — far from the cluster in $x$ and far from the line in $y$ — is the most influential.

Example 2

Most points sit in

x \in [0, 10]

. Classify each new point. (i) A point at

x = 50

that lies on the trend line. (ii) A point at

x = 5, y = 100

when the trend predicts

y \approx 20

x = 5

. (iii) A point at

x = 50, y = 100

when the trend predicts

y \approx 30

x = 50

. Which classification is correct? (A) (i) influential; (ii) leverage; (iii) y-outlier (B) (i) high leverage but low influence; (ii) y-outlier; (iii) influential (both) (C) (i) y-outlier; (ii) influential; (iii) leverage (D) (i) and (iii) both have no effect on the fit.

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Type 3

Justifying removal

Removing a point requires a documented reason — a recording error or a different-population argument. 'Without it $r$ is higher' is not a reason.

Example 3

A student removes the most extreme outlier from a dataset because

r

rises from 0.6 to 0.85 after the drop, and reports only the cleaned data. Which critique is correct? (A) The student is correct because the goal of regression is to maximize

r

. (B) The removal is data manipulation. Defensible removals require a recording-error or different-population justification; otherwise report the analysis both with and without the point. (C) Outliers should never be removed under any circumstance. (D) Removal is fine only if it improves

R^2

by more than 30 percentage points.

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